Question: Factor the following expression: $x^2 + x - 6$
Solution: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & +& {1}x& & {-6} \end{eqnarray} $ The coefficient on the $x$ term is $1$ and the constant term is $-6$ , so to reverse the steps above, we need to find two numbers that add up to $1$ and multiply to $-6$ You can try out different factors of $-6$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {1}$ $ {a} \times {b} = {-6}$ The two numbers $3$ and $-2$ satisfy both conditions: $ {3} + {-2} = {1} $ $ {3} \times {-2} = {-6} $ So we can factor the expression as: $(x + {3})(x {-2})$